Gamma Distribution and Hypothesis Testing

The problem asks us to find the rejection region for a Uniformly Most Powerful (UMP) test for the parameter λ of a gamma distribution. We are given a random sample \(X_1, X_2, \dots, X_6\) from a gamma distribution with probability density function \(f(x \mid \lambda)=\frac{\lambda^4}{6} e^{-\lambda x} x^3\) for \(x>0\), where \(\lambda > 0\). We need to test the null hypothesis \(H_0 : \lambda = 1\) against the alternative hypothesis \(H_1 : \lambda > 1\) at a significance level of \(\alpha = 0.05\).

UMP Test for Exponential Family

The given probability density function can be recognized as a gamma distribution with shape parameter \(k=4\) and rate parameter \(\lambda\). The general form of the Gamma(k, \(\lambda\)) pdf is \(f(x; k, \lambda) = \frac{\lambda^k}{\Gamma(k)} x^{k-1} e^{-\lambda x}\). Comparing this with the given pdf, we see that \(k-1=3 \implies k=4\), \(\lambda\) is the rate parameter, and \(\Gamma(4)=3!=6\). So, \(X_i \sim \text{Gamma}(4, \lambda)\).

The pdf can be written in the exponential family form: \(f(x \mid \lambda) = c(\lambda) h(x) e^{w(\lambda) t(x)}\), where:

• \(c(\lambda) = \frac{\lambda^4}{6}\)
• \(h(x) = x^3\)
• \(w(\lambda) = -\lambda\)
• \(t(x) = x\)

The parameter space is \(\Lambda = (0, \infty)\). We are testing \(H_0: \lambda = 1\) against \(H_1: \lambda > 1\). This is a one-sided test.

For distributions in the exponential family, the sum of the \(t(X_i)\) values, \(T = \sum_{i=1}^n t(X_i)\), is a sufficient statistic for the parameter. In this case, \(T = \sum_{i=1}^6 X_i\).

According to the theory of UMP tests for the exponential family, for a one-sided alternative like \(H_1: \lambda > \lambda_0\), the UMP test depends on whether \(w(\lambda)\) is increasing or decreasing in \(\lambda\).

• If \(w(\lambda)\) is strictly increasing, the UMP test rejects \(H_0\) for large values of \(T = \sum t(X_i)\).
• If \(w(\lambda)\) is strictly decreasing, the UMP test rejects \(H_0\) for small values of \(T = \sum t(X_i)\).

In our case, \(w(\lambda) = -\lambda\), which is a strictly decreasing function of \(\lambda\). The alternative hypothesis is \(H_1: \lambda > 1\). Therefore, the UMP test of size \(\alpha\) for testing \(H_0: \lambda = 1\) against \(H_1: \lambda > 1\) rejects \(H_0\) if and only if \(T = \sum_{i=1}^6 X_i \leq c\) for some critical value \(c\).

Distribution of the Test Statistic T under H0

Under the null hypothesis \(H_0: \lambda = 1\), each \(X_i\) is independently distributed as Gamma(4, 1). The sum of independent gamma random variables with the same rate parameter \(\lambda\) is also a gamma random variable with shape parameter equal to the sum of individual shape parameters and the same rate parameter.

Given \(X_i \sim \text{Gamma}(k_i, \lambda)\) for \(i=1, \dots, n\), the sum \(Y = \sum_{i=1}^n X_i \sim \text{Gamma}(\sum_{i=1}^n k_i, \lambda)\).

Here, \(n=6\), and for each \(i\), \(k_i=4\) and under \(H_0\), \(\lambda=1\). Thus, the test statistic \(T = \sum_{i=1}^6 X_i\) follows a gamma distribution:

\(T \sim \text{Gamma}(\sum_{i=1}^6 4, 1) = \text{Gamma}(6 \times 4, 1) = \text{Gamma}(24, 1)\) under \(H_0\).

Relating Gamma Distribution to Chi-squared Distribution

There is a known relationship between the gamma distribution and the chi-squared distribution. If a random variable \(Y\) follows a Gamma(k, \(\lambda\)) distribution, then the random variable \(2 \lambda Y\) follows a chi-squared distribution with \(2k\) degrees of freedom, i.e., \(2 \lambda Y \sim \chi^2_{2k}\).

In our case, under \(H_0\), \(T \sim \text{Gamma}(24, 1)\). Using the relationship, with \(Y=T\), \(k=24\), and \(\lambda=1\), we have \(2 \times 1 \times T = 2T \sim \chi^2_{2 \times 24} = \chi^2_{48}\) under \(H_0\).

Determining the Critical Value

The rejection region for the UMP test is \(T \leq c\). The size of the test is given as \(\alpha = 0.05\). The size of the test is the probability of rejecting \(H_0\) when \(H_0\) is true.

\(P(\text{Reject } H_0 \mid H_0) = P(T \leq c \mid \lambda = 1) = 0.05\).

Since \(2T \sim \chi^2_{48}\) under \(H_0\), we can rewrite the probability in terms of \(2T\):

\(P(2T \leq 2c \mid \lambda = 1) = 0.05\).

Let \(Y = 2T\). Then \(Y \sim \chi^2_{48}\) under \(H_0\). We need to find \(c\) such that \(P(Y \leq 2c) = 0.05\).

The question defines \(\chi^2_{v, \alpha}\) as the \((1-\alpha)\)th quantile of the \(\chi^2_v\) distribution. This means \(P(\chi^2_v \leq \chi^2_{v, \alpha}) = 1 - \alpha\).

We need the value \(x\) such that \(P(\chi^2_{48} \leq x) = 0.05\). This value is the 0.05th quantile of the \(\chi^2_{48}\) distribution.

Using the notation \(\chi^2_{v, \alpha}\) for the \((1-\alpha)\)th quantile, the 0.05th quantile is found by setting \(1-\alpha = 0.05\), which means \(\alpha = 0.95\). Thus, the 0.05th quantile of \(\chi^2_{48}\) is \(\chi^2_{48, 0.95}\).

Therefore, we must have \(2c = \chi^2_{48, 0.95}\).

Solving for \(c\), we get \(c = \frac{1}{2} \chi^2_{48, 0.95}\).

The rejection region is \(T \leq c\), which means \(T \leq \frac{1}{2} \chi^2_{48, 0.95}\).

Comparing with Options

Let's compare our derived rejection region with the given options:

1. \(T \geq \frac{1}{2} \chi_{48,0.05}^2\) - This is a rejection region for large T and uses the 95th quantile (\(1-0.05=0.95\)th quantile is denoted \(\chi^2_{48, 0.05}\)). Our test rejects for small T.
2. \(T \leq \frac{1}{2} \chi_{48,0.95}^2\) - This is a rejection region for small T and uses the 0.05th quantile (\(1-0.95=0.05\)th quantile is denoted \(\chi^2_{48, 0.95}\)). This matches our result.
3. \(T \geq \frac{1}{2} \chi_{24,0.05}^2\) - Incorrect degrees of freedom (24 instead of 48) and wrong rejection direction.
4. \(T \leq \frac{1}{2} \chi_{24,0.95}^2\) - Incorrect degrees of freedom (24 instead of 48).

Option 2 correctly represents the rejection region.

The test ψ rejects \(H_0\) if and only if \(T \leq \frac{1}{2} \chi_{48,0.95}^2\).